It is better to fill out the part at slower speeds in-order to let any compressed air escape, and to get the best surface finish on the part.3 • Using all of the previous settings, record the Transfer Pressure, Fill Time, and Injection speed • Reduce the injection velocity by 10% and record the same values • Keep reducing the injection velocity until a fill time of 10-15 seconds is reached Transfer Pressure (psi) Fill Time (sec) Injection Velocity (in/sec) Relative Shear Rate Relative Viscosity 1238 0 .20 8 .00 5 .00 247 .60 1125 0 .22 7 .00 4 .55 247 .50 1058 0 .25 6 .00 4 .00 264 .50 960 0 .35 5 .00 2 .86 336 .12 870 0 .45 4 .00 2 .22 391 .65 780 0 .58 3 .00 1 .72 452 .59 690 0 .85 2 .00 1 .18 586 .78 600 1 .35 1 .00 0 .74 810 .45 510 2 .50 0 .75 0 .40 1275 .83 420 5 .23 0 .50 0 .19 2198 .34 330 9 .56 0 .25 0 .10 3157 .99 240 15 .26 0 .10 0 .07 3667 .49 FIGURE 47 - Injection Speed 22 Gravi-Tech • Calculate the Relative Viscosity for each of the Injection Speeds using the equation above (Transfer Pressure * Fill Time) • Calculate the Relative Shear Rate for each of the injection speeds using the equation (1/Fill Time) • Plot the points on a graph and find the optimum velocity • The optimum velocity will be where the curve starts to level out In figure 48 above, the optimum velocity is where the plotted line crosses the black trend line; therefore the optimum velocity would be about 1 .00 in/sec . FIGURE 48 - The Injection Unit Nozzle Non-Return Valve Heater Bands Barrel Screw Hopper Metering Zone Compression Zone Feed Zone FIGURE 47 - Non-Return Valve: Check Ring Check Ring Open Check Ring Closed FIGURE 49 - The Injection Unit FIGURE 50 - Non-return Valve: Check Ring Design Guide 45 FIGURE 48 - Non-Return Valve: Ball Check Ball Check Open Ball Check Closed Poppet Valve The final type of non-return valve is called a poppet valve . End of Fill Part Length Dynamic Pressure Hydrostatic Pressure P re s s u re Gate End Part FIGURE 61 - Deflection Equations H F WLMax Deflection: 0.002" (0.05mm) 1 = W • H3 12 _______ bending = F • L3 48 • E • I _______ 4 π tc = h2 1n π2 • a • Tmelt – Tcoolant Teject – Tcoolant tc = D2 1.61n 23.1 • a Tmelt – Tcoolant Teject – Tcoolant a = k p * Cp Qmoldings = mmoldings • Cp • Tme • Cplt – Teject cooling nlines moldings tccooling Vcoolant line nmax, coolant • Pcoolant • Cp, coolant Dmax = 4 • Pcoolant • Vcoolant π • µcoolant • 4000 Dmin = Pcoolant • Lline • V2coolant5 10π • ∆Pline 4 π tc = h2 1n π2 • a • Tmelt – Tcoolant Teject – Tcoolant tc = D2 1.61n 23.1 • a Tmelt – Tcoolant Teject – Tcoolant a = k p * Cp Qmoldings = mmoldings • Cp • Tme • Cplt – Teject cooling nlines moldings tccooling Vcoolant line nmax, coolant • Pcoolant • Cp, coolant Dmax = 4 • Pcoolant • Vcoolant π • µcoolant • 4000 Dmin = Pcoolant • Lline • V2coolant5 10π • ∆Pline 4 π tc = h2 1n π2 • a • Tmelt – Tcoolant Teject – Tcoolant tc = D2 1.61n 23.1 • a Tmelt – Tcoolant Teject – Tcoolant a = k p * Cp Qmoldings = mmoldings • Cp • Tme • Cplt – Teject cooling nlines moldings tccooling Vcoolant line nmax, coolant • Pcoolant • Cp, coolant Dmax = 4 • Pcoolant • Vcoolant π • µcoolant • 4000 Dmin = Pcoolant • Lline • V2coolant5 10π • ∆Pline FIGURE 60 - Pressure vs Part Length FIGURE 61 - Deflection equations FIGURE 62 - For Plate Shaped Parts FIGURE 63 - For Cylindrical Shaped Parts Design Guide 49 • M Moldings = Combined mass of molded parts • C p = Specific Heat of the material Step 3 – Heat Removal Rate • N lines = The total number of independent cooling lines there are in the mold • t c = The cooling time required by the part (Determined in step 1) Step 4 – Coolant Volumetric Flow Rate • ΔT Max,Coolant = Change in coolant Temperature During Molding (1°C) • ρ Coolant = Density of coolant • CP = Specific heat of coolant Step 5 – Determine Cooling Line Diameter • ρ Coolant = Density of coolant • V Coolant = Volumetric flow rate of coolant • μ Coolant = Viscosity of coolant • ΔP line = Max pressure drop per line (Usually equals half of the pump capacity) • L Line = Length of the cooling lines COOLING LINE SPACING 4 π tc = h2 1n π2 • a • Tmelt – Tcoolant Teject – Tcoolant tc = D2 1.61n 23.1 • a Tmelt – Tcoolant Teject – Tcoolant a = k p * Cp Qmoldings = mmoldings • Cp • Tme • Cplt – Teject cooling nlines moldings tccooling Vcoolant line nmax, coolant • Pcoolant • Cp, coolant Dmax = 4 • Pcoolant • Vcoolant π • µcoolant • 4000 Dmin = Pcoolant • Lline • V2coolant5 10π • ∆Pline 4 π tc = h2 1n π2 • a • Tmelt – Tcoolant Teject – Tcoolant tc = D2 1.61n 23.1 • a Tmelt – Tcoolant Teject – Tcoolant a = k p * Cp Qmoldings = mmoldings • Cp • Tme • Cplt – Teject cooling nlines moldings tccooling Vcoolant line nmax, coolant • Pcoolant • Cp, coolant Dmax = 4 • Pcoolant • Vcoolant π • µcoolant • 4000 Dmin = Pcoolant • Lline • V2coolant5 10π • ∆Pline 4 π tc = h2 1n π2 • a • Tmelt – Tcoolant Teject – Tcoolant tc = D2 1.61n 23.1 • a Tmelt – Tcoolant Teject – Tcoolant a = k p * Cp Qmoldings = mmoldings • Cp • Tme • Cplt – Teject cooling nlines moldings tccooling Vcoolant line nmax, coolant • Pcoolant • Cp, coolant Dmax = 4 • Pcoolant • Vcoolant π • µcoolant • 4000 Dmin = Pcoolant • Lline • V2coolant5 10π • ∆Pline 4 π tc = h2 1n π2 • a • Tmelt – Tcoolant Teject – Tcoolant tc = D2 1.61n 23.1 • a Tmelt – Tcoolant Teject – Tcoolant a = k p * Cp Qmoldings = mmoldings • Cp • Tme • Cplt – Teject cooling nlines moldings tccooling Vcoolant line nmax, coolant • Pcoolant • Cp, coolant Dmax = 4 • Pcoolant • Vcoolant π • µcoolant • 4000 Dmin = Pcoolant • Lline • V2coolant5 10π • ∆Pline 4 π tc = h2 1n π2 • a • Tmelt – Tcoolant Teject – Tcoolant tc = D2 1.61n 23.1 • a Tmelt – Tcoolant Teject – Tcoolant a = k p * Cp Qmoldings = mmoldings • Cp • Tme • Cplt – Teject cooling nlines moldings tccooling Vcoolant line nmax, coolant • Pcoolant • Cp, coolant Dmax = 4 • Pcoolant • Vcoolant π • µcoolant • 4000 Dmin = Pcoolant • Lline • V2coolant5 10π • ∆Pline 4 π tc = h2 1n π2 • a • Tmelt – Tcoolant Teject – Tcoolant tc = D2 1.61n 23.1 • a Tmelt – Tcoolant Teject – Tcoolant a = k p * Cp Qmoldings = mmoldings • Cp • Tme • Cplt – Teject cooling nlines moldings tccooling Vcoolant line nmax, coolant • Pcoolant • Cp, coolant Dmax = 4 • Pcoolant • Vcoolant π • µcoolant • 4000 Dmin = Pcoolant • Lline • V2coolant5 10π • ∆Pline 2D < H line < 5D H line < W line < 2H line FIGURE 70 - Cooling Line Spacing FIGURE 64 - Heat Transfer Equation FIGURE 65 - Total Cooling for Mold FIGURE 66 - Cooling Required by Each Line FIGURE 68 - Max Diameter Equation FIGURE 69 - Min Diameter Equation FIGURE 67 - Volumetric Flow Rate Equation 50 Gravi-Tech ADHESIVE ADVANTAGES DISADVANTAGES Cyanoacrylate Rapid, one-part process Various viscosities Can be paired with primers for polyolefins Poor strength Low stress crack resistance Low chemical resistance Epoxy High strength Compatible with various substrates Tough Requires mixing Long cure time Limited pot life Exothermic Hot Melt Solvent-free High adhesion Different chemistries for different substrates High temp dispensing Poor high temp performance Poor metal adhesion Light Curing Acrylic Quick curing One component Good environmental resistance Oxygen sensitive Light source required Limited curing configurations Polyurethane High cohesive strength Impact and abrasion resistance Poor high heat performance Requires mixing Silicone Room temp curing Good adhesion Flexible Performs well in high temps Low cohesive strength Limited curing depth Solvent sensitive No-Mix Acrylic Good peel strength Fast cure Adhesion to variety of substrates Strong odor Exothermic Limited cure depth Design Guide 51 Bibliography 1 .

Feb 10, 2025

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